SOLUTION: Solve for x over the complex numbers: y=x^3+4x^2; y=3x+18 thanks a bunch

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Solve for x over the complex numbers: y=x^3+4x^2; y=3x+18 thanks a bunch      Log On


   

Question 939079: Solve for x over the complex numbers: y=x^3+4x^2; y=3x+18
thanks a bunch
Found 2 solutions by Alan3354, ewatrrr:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
y=x^3+4x^2; y=3x+18
y=x^3+4x^2 = 3x+18
f(x) = x^3 + 4x^2 - 3x - 18 = 0
-------------
Try the factors of 18, + and minus.
eg, f(1) = 1 + 4 - 3 - 18 = -16 Not a zero.
Find any real integer zeroes then divide by (x - zero). One zero will reduce it to a quadratic.

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
y=x^3+4x^2; y=3x+18
...
x^3+4x^2 = 3x+18
x^3 + 4x^2 - 3x - 18 = 0
x = 2 a solution 8%2B16-6-18+=+0
...
Using Synthetic Div1sion
 

2	1	4	-3	-18

		2	12	18

	1	6	9	0

x^3 + 4x^2 - 3x - 18 = (x-2)(x^2 + 6x + 9)

will Let You finish it Up